Staff: Mentor

The curve for every k gets closer and closer to a normal distribution with the same mean and variance with increasing k.
If you scale the distribution in an appropriate way, you get something approaching a normal distribution with mean 0 and variance 1.

Putting k=μ (mean of the normals, I presume) appears weired, k is positive integer ( being the number of normals summed here), and -< μ< ∞ is real. Also that, if all means of the initial normal distributions are not 0, the then the resulting chi sq is non central.

Staff: Mentor

Where is the problem in different gaussian distributions which all have an integer as expectation value?
The chi-squared distribution is positive for positive values only, but for large k, the gaussian distribution is a reasonable approximation (its part <0 is negligible).

Please check again. I am talking of μ as normal mean.... you are mistaking μ as chi sq mean. "μ =k" CAN NOT be consequence of any literature definition, where ever written....lodge a request for correction there. And of course, I stand correct about non centrality.... please go through the derivation of n.c. chi sq.

Where is the problem in different gaussian distributions which all have an integer as expectation value?
The chi-squared distribution is positive for positive values only, but for large k, the gaussian distribution is a reasonable approximation (its part <0 is negligible).

About integer and real part: I did not say that a particular value of normal mean cannot be integer. But I say, taking normal mean as integer is weired. The first loophole arises in context of the present problem as the fact that μ is differentiable but k is not.

I am talking of μ as normal mean.... you are mistaking μ as chi sq mean.

In that case, I am not sure of your question, because you referred to the original μ=k, and in the original context, μ is the mean of the chi squared distribution.

please go through the derivation of n.c. chi sq.

I'm also not sure whether this is a suggestion for me to go through it myself, or to write down the derivation here in this post. In the latter case, probably another contributor would do a better job of it than I would.

Well, if μ is assumed as chi sq mean, no issues (is it not obvious from my posts). The original post is some what misleading with (unnecessary) involvement of μ as the chi sq mean... where k clearly stands for that. Without clarification, μ has been naturally presumed as the originating normal mean. I understood your problem in a completely wrong way altogether.
Hope it clarifies my statements.
PS. "going through" in common jargon probably does not mean writing down. :)

All's well that ends well. That's what I like about mathematics (and mathematicians): if people talk at cross purposes, it quickly gets cleared up. Unlike in most disciplines. So I guess this thread can be closed.